## College Physics (4th Edition)

The distance between the source and the detector is $80~km$
We can use the speed $v_1$ and the travel time $t_1$ of the longitudinal waves to write an expression for the distance: $d = v_1~t_1$ Note that $t_2 = t_1+2.0~s$. We can use the speed $v_2$ and the travel time $t_2$ of the transverse waves to write an expression for the distance: $d = v_2~t_2 = v_2~(t_1+2.0~s)$ Since the distance is the same, we can equate the two equations to find $t_1$: $v_1~t_1 = v_2~(t_1+2.0~s)$ $(v_1-v_2)~t_1 = (2.0~s)~v_2$ $t_1 = \frac{(2.0~s)~v_2}{v_1-v_2}$ $t_1 = \frac{(2.0~s)(8.0~km/s)}{10.0~km/s-8.0~km/s}$ $t_1 = 8.0~s$ We can find the distance $d$: $d = v_1~t_1 = (10.0~km/s)(8.0~s) = 80~km$ The distance between the source and the detector is $80~km$.